So in the first one, we can figure out the period of the function based on how long it takes to repeat.
I guess the easiest point to look at would just be where it crosses the x-axis.
So it crosses at \(x=6\pi\) then again at \(x=7\pi\).
So that function would have a period of \(\pi\).
Hmm you'll want to remember some of your basic trig graphs.
I think this one lines up with the tangent function.
But ummmm let's think about what's happening by the origin so we can make sure it lines up correctly.
Tangent goes through the origin. Would this function as well? :o

Yah it looks like it. It'll go through \(5\pi\), \(4\pi\), \(3\pi\), \(2\pi\), \(\pi\) and \(0\pi\).
I'm pretty sure the first graph is just \(y=\tan x\).
It doesn't appear to have any transformations applied to it. Hmmmm D:

|dw:1358117529784:dw|This first one is kinda tricky. I think they wanted you to try and figure out which function it is without knowing exactly what it looks like about the origin.
If you know what it looks like about the origin you can probably just match it up with one of the trig graphs that you remember.
So in this first one, they want you to notice the SHAPE that it has (it still is the tangent function based on it's shape) and label any transformations that have been applied to it (shifts to the right, left, up, down, or scaling).
Hmmmmmmmm let's check out this second one a sec :O

Yah it has that same shape doesn't it? Hmm
But we have a small problem, instead of passing through the origin, the function has an asymptote at the origin, which means it has been altered a little bit.
It'll still end up being \(y=\tan x\), but with some alteration applied to it :D Let's see if we can figure it out.

Remember how to integrate the period into the function?\[\large y=\tan \color{#F35633}{b}x\]The period of this function is, \(\large \frac{\pi}{\color{#F35633}{b}}\)
Andddd we determined that the period is \(2\pi\),
Solving for b gives us,\[\large \frac{\pi}{\color{#F35633}{b}}=2\pi \qquad \rightarrow \qquad \color{#F35633}{b}=\frac{1}{2}\]

Wellllll it looks like it also has a shift applied to it, shifting it to the right (or left I suppose..). Instead of passing through the origin it's passing through \(\pi\), so it has been shifted to the right \(\pi\).
\[\large y=\tan\left[\frac{1}{2}\left(x-\pi\right)\right]\]I think this is the answer we're trying to get. Ugh I'm not 100% confident in that. I wish we had an answer key to check it against :D